If the quantum numbers of an ion are given, how can we identify this ion by its quantum numbers?

For example :

n = 3; $l$ = 2; $m_l$ = 0; m$_s$ = + ½

What I have so far: $n$ is 3 and $l$ is 2 which corresponds to $d$, so the last shell is $3d1$, therefore we have 21 $e^-$ and the answer will be $\ce{Sc}$ ion (from the periodic table) and the answer I got is $\ce{Mn^{2+}}$. Is this the correct approach?

  • 2
    $\begingroup$ Please show how you arrived at Mn+. As I mentioned on your other question, it's critical that you put forth more effort towards these problems. $\endgroup$
    – jonsca
    Jul 10 '14 at 16:32
  • $\begingroup$ Why !! at the end of each chapter in the book there are a lot of problems .. some of them, their answers are at the end of the book .. but only the final answer is there .. So, I took one of the solved questions that I dont understand and I put it here because i have no idea how to start solving it.. consequently, how do you expect me to show some effort about something that I don't know the way it is solved ? $\endgroup$
    – Maher
    Jul 10 '14 at 20:13
  • $\begingroup$ it is better to know : I am not scientist and my major has nothing to do with chemistry, but it is a general course that I have to take .. it is hard enough.. don't make it more ! at least write down some guidelines that I should follow to show some effort :( $\endgroup$
    – Maher
    Jul 10 '14 at 20:16
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    $\begingroup$ Maher - start by explaining what you do understand. You must know something - even if it doesn't seem like it will help. That way we have a better idea of where to start. For example, think about the quantum numbers - what do they refer to? Is it a whole atom? All the electrons? Some of the electrons? If we know what you do understand, it is a lot easier to fill in the gaps. Otherwise, we are left with trying to give an introductory quantum chemistry lecture in the space of one answer. $\endgroup$
    – thomij
    Jul 10 '14 at 20:37
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    $\begingroup$ I have given you the proper framework for what we expect from those asking questions based on the information given in your comment. Please follow this from now on. $\endgroup$
    – jonsca
    Jul 10 '14 at 22:19

A pattern-based approach: Remember that the values for ${m_l}$ are centered about 0. This means for a d-orbital, where ${l=2}$, the values of ${m_l}$ = ${(-2, -1, 0, +1, +2)}$.

So, for ${l=2}$ and ${m_l=0}$ and ${m_s=+1/2}$, you should arrive at ${d^3}$. The orbitals each get an electron for all the positive spins first, then are filled completely with the negative spin ones.

That means you "visit" ${m_l}$ twice (in rising order: -2, -1, 0, +1, +2, repeat).

This means the first d-electron (↑ _ _ _ _) such as ${Sc}$ has the configuration ${m_l=-2, m_s=+1/2.}$

The 3rd electron (↑ ↑ ↑ _ _) would have ${m_l=0, m_s=+1/2}$.

A 5-electron configuration (↑ ↑ ↑ ↑ ↑) would have ${m_l=+2, m_s=+1/2}$.

Now on the right-hand side of the d-group, the 6th electron (↑↓ ↑ ↑ ↑ ↑) would have ${m_l=-2, m_s=-1/2}$, and then the magnetic quantum numbers rise to +2 again at ${d10}$.

Anyways, ${Mn^{+2}}$ is ${d^{3}}$ (5 to start, then subtract 2 because of its +2 charge) which matches the quantum numbers you specified.

It's not an 8 electron configuration because of the spin specified. That'd be (↑↓ ↑↓ ↑↓ ↑ ↑).


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